# If a is odd, prove ... by division algorithm

• Sep 9th 2012, 09:52 AM
janchilds
If a is odd, prove ... by division algorithm
The problem states that if a is odd, then prove 12 | a^2 + (a+2)^2 + (a+4)^2 + 1

Here is what I did but it seems more by induction than the division algorithm. If someone could let me know if I am on the right track.

Base case a=1 .... 12 |36 done.

So, if k is in G then prove (2k-1) is in G

12 | (2k-1)^2 + ((2k-1)+2)^2 + ((2k-1)+4)^2 +1 - I simply multiplied this whole thing out and got 12 | 12k^2 + 12k + 12

Then if a|b we know a|bc, so 12|12 or 12|-12 if we want to use if a|b and a|c then a|b-c

Thus k+1 is in G.

I guess I just thought I would be looking for k(k+1)(k+2) as we did in the example in class.
• Sep 9th 2012, 10:16 AM
MaxJasper
Re: If a is odd, prove ... by division algorithm
a=odd = $a\text{=}2k+1$

$a^2+(a+2)^2+(a+4)^2+1 = 12 \left(3+3 k+k^2\right)$
• Sep 9th 2012, 10:33 AM
Soroban
Re: If a is odd, prove ... by division algorithm
Hello, janchilds!

You are trying an inductive proof . . . Okay!

Quote:

$\text{If }a\text{ is odd, then }a^2 + (a+2)^2 + (a+4)^2 + 1\text{ is divisible by 12.}$

Verify $S(1)\!:\;1^2 + 3^2 + 5^2 + 1 \:=\:36\;\hdots\;\text{True!}$

Assume $S(k)\!:\;k^2 + (k+2)^2 +(k+4)^2 + 1 \;=\;12p\;\text{ for some integer }p.$

Add $(k+6)^2-k^2$ to both sides:

. . $(k+2)^2 + (k+4)^2 + (k+6)^2 + 1 \;=\;12p + (k+6)^2-k^2$

. . $(k+2)^2 + (k+4)^2 + (k+6)^2 + 1 \;=\;12p + k^2 +12k + 36 - k^2$

. . $(k+2)^2 + (k+4)^2 + (k+6)^2 + 1 \;=\;12p + k + 36$

. . $(k+2)^2 + (k+4)^2 + (k+6)^2 + 1 \;=\;\underbrace{12(p + k + 3)}_{\text{divisible by 12}}$

We have proved $S(k\!+\!1).$
The inductive proof is complete.
• Sep 9th 2012, 12:55 PM
janchilds
Re: If a is odd, prove ... by division algorithm
I get this... a bit different format from what I was doing but the same bottom line.

But this is proving it by Induction. If the instructions specifically says prove using the Division Algorithm.... is that the same?

Thanks,
Janet
• Sep 9th 2012, 01:14 PM
SlipEternal
Re: If a is odd, prove ... by division algorithm
You can add the division algorithm into the proof rather easily. Write the statement of the proof, then any time you are discussing divisibility, you just need $r=0$ where $r$ is the remainder in the division algorithm. So, your proof by induction changes form. You claim that for any $a\in 2\mathbb{Z}+1$, there exists $p\in \mathbb{Z}$ so that $(a^2+(a+2)^2+(a+4)^2+1) = 12p + 0$. Now, you are using the division algorithm where your $q=12$ and $r=0$.